a, \(n_{Br_2}=0,15.1=0,15\left(mol\right)\)
PT: \(C_2H_4+Br_2\rightarrow C_2H_4Br_2\)
Theo PT: \(n_{C_2H_4}=n_{Br_2}=0,15\left(mol\right)\Rightarrow\%V_{C_2H_4}=\dfrac{0,15.22,4}{8,96}.100\%=37,5\%\)
\(\Rightarrow\%V_{CH_4}=62,5\%\)
b, \(CH_4+2O_2\underrightarrow{t^o}CO_2+2H_2O\)
\(C_2H_4+3O_2\underrightarrow{t^o}2CO_2+2H_2O\)
Ta có: \(n_{CH_4}=\dfrac{8,96.62,5\%}{22,4}=0,25\left(mol\right)\)
Theo PT: \(n_{CO_2}=n_{CH_4}+2n_{C_2H_4}=0,55\left(mol\right)\)
PT: \(CO_2+Ca\left(OH\right)_2\rightarrow CaCO_{3\downarrow}+H_2O\)
\(n_{CaCO_3}=n_{CO_2}=0,55\left(mol\right)\Rightarrow m_{CaCO_3}=0,55.100=55\left(g\right)\)
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